# Sheep in wolf's clothing

In the following exhibits we give an advanced or alternative way of thinking about mathematics concepts which are likely to be known in a more familiar form.

Explore these structures and experiment by substituting particular values such as $0, \pm 1$. Can you work out what they represent?

Exhibit A

All pairs of integers such that:

$$(a, b) + (c, d) = (ad+bc, bd)\quad\quad (Na, Nb) \equiv (a, b) \mbox{ for all } N\neq 0$$

Can you find two pairs which add up to give $(0, N)$ or $(0, M)$ for various values of $N$, $M$?

A set of ordered pairs of real numbers which can be added and multiplied such that

$(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 +y_2)$

$(x_1, y_1)\times (x_2, y_2) = (x_1x_2 -y_1y_2, x_1y_2+y_1x_2)$

Exhibit C

A set defined recursively such that

$+_k(1) = +_1(k)$

$+_k(+_1(n)) = +_1(+_k(n))$

$\times_k(1) = k$

$\times_k(+_1(n)) = +_k(\times_k(n))$

In these rules, $k$ and $n$ are allowed to be any natural numbers

Once you have figured out what these structures represent ask yourself this: Are these good representations? What benefits can you see to such a representation? How might familiar properties from the structures be represented in these ways?

**Exhibit A**

The condition $$(Na, Nb) \equiv (a, b) \mbox{ for all } N\neq 0$$ ought to give it away: $$(a,b) \iff \frac{a}{b}$$ This statement simply says that if the numerator and denominator of a fraction share a common factor, they can be cancelled down.

**Exhibit B**

If we represent a complex number a+bi by the ordered pair (a,b), we get the required properties:

$$(a+bi) + (c+di) = (a+c) + (b+d)i \iff (a,b) + (c,d) = (a+c, b+d)$$

$$(a+bi) \times (c+di) = (ac-bd) + (ad+bc)i \iff (a,b) \times (c,d) = (ac-bd, ad+bc)$$

**Exhibit C**

These formally define addition and multiplication over the natural numbers. Can you see how the familiar properties we're used to follow from them?

The first implies $k+1 = 1+k$, i.e. addition is commutative.

The second implies $k+(1+n) = 1+(k+n)$, i.e. addition is associative.

The third implies $k\times 1 = k$, i.e. 1 is the multiplicative identity.

The fourth implies $k\times(1+n) = k+(k\times n)$, that mulitplication is distributative over addition.

This is a rigorous treatment of a very familiar concept. For more information on this subject, you could start by reading this Wikipedia article.